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Finding the transfer Function of a Voltage Buffer

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  • #1
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This is the circuit that I'm trying to work with.

Essentially, as far as I can see it's a voltage divider connected to a voltage follower. I have the output of the first op amp as: H(s)=S1/In1= 1/10 = 0.1.

However I'm not sure if there are how to obtain characteristics of such a simple transfer function. When I put it in wolfram mathematica to see the Bode Plot; it says something about it not being a real transfer function (works fine when I put a random s in there).

My guess for the circuit is that the gain is always 0.1, and that there should be no phase shift. Can anyone say if I am missing anything? Or point me in the direction of obtaining a mathematical proof?

Also what is the significance of the resistor that is said to be added for bias reasons. I'm finding it difficult to find resources that can explain it to me.

Thanks,
M
 

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  • #2
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I haven't looked at the circuit closely, but seeing that there are no capacitors or inductors, I do believe that your equation seems correct. If there are no capacitors or inductors, your transfer function should just be a constant. (Of course, this is mainly because your circuit is very idealized).

As for the bias resistor, I'm guessing this resistor is just there to ground the +terminal. As for why they picked 10k, I'm thinking that is based on the input impedance of the op amp, and therefore will set the input current into the op amp. I'm guessing the value actually makes no difference in the problem since you are probably assuming an op amp with an infinite input impedance.

I'm not sure what the deal is with the arrow pointing into the +terminal of the op amp. I think that is just there to indicate that the input current to the op amp is not 0 and the input impedance is not ∞. (But I imagine that the input impedance should still be very high, and the input current close to 0, or at least on the order of uA)
 
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  • #3
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Also, since the op amps are idealized, and there is no capacitance or inductance, when your transfer function is plotted in the frequency domain, it will just be a constant .1 for all frequencies, which is probably why wolfram mathematica had a problem with it. (Remember, this is only because your op amps are idealized).
 
  • #4
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Thanks a lot, we're actually just beginning to scratch that unbearable uncertaintly of non-idealised op amps. And considering this is a something will be be demostrating in the lab (and considering you repeatedly said only for ideal op amps); I'm wondering what you think the effect will be with an oscilloscope comparing the input and the output.

Also are you saying that I'm correct in guessing there would be no phase shift for an ideal op amp?

Thanks a lot : )
 
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  • #6
rude man
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This is the circuit that I'm trying to work with.

Also what is the significance of the resistor that is said to be added for bias reasons. I'm finding it difficult to find resources that can explain it to me.

Thanks,
M
The 10K resistor does absolutely nothing. The ouput impdance of the preceding amplifier provides all the biasing needed, and its output impedance is much, much lower than 10K (it's zero with ideal op amps).

As for the rest, yes you have the right gain.
 
  • #7
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Thanks a lot, we're actually just beginning to scratch that unbearable uncertaintly of non-idealised op amps. And considering this is a something will be be demostrating in the lab (and considering you repeatedly said only for ideal op amps); I'm wondering what you think the effect will be with an oscilloscope comparing the input and the output.

Also are you saying that I'm correct in guessing there would be no phase shift for an ideal op amp?

Thanks a lot : )
Yes, there should be no phase shift for an ideal op amp. Even in the lab, I imagine you won't be able to see a phase shift between the input and the output unless you zoom in extremely close.

I'm not sure what the effect of the bias resistor will be in the lab. It depends on the op amp you're using, and the input impedance and the input current requirements of the op amp. But I imagine that you will find it not working correctly if that resistance is too high or too low.
 
  • #8
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The 10K resistor does absolutely nothing. The ouput impdance of the preceding amplifier provides all the biasing needed, and its output impedance is much, much lower than 10K (it's zero with ideal op amps).

As for the rest, yes you have the right gain.
I could see that. Its probably just there in case the output of the first op amp becomes disconnected, you don't have a floating input on the second op amp.
 
  • #9
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Thanks a lot to both of you for the information that you have provided! I should be good now, if I have any more questions I'll ask my lab instructor : )
 

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